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Differential equation

  1. May 6, 2007 #1
    1. The problem statement, all variables and given/known data

    I am to show that the differential equataion D^2f+[A+V(x)]f=0
    {A is a constant and V(x+m)=V(x)}
    has the solutions of the form f=exp[ikx]U(x)where U(x+m)=U(X)


    2. Relevant equations



    3. The attempt at a solution

    I tried to differentiate the given solution and put it in the equation...but the method is not working.Can you please help?
     
  2. jcsd
  3. May 7, 2007 #2

    siddharth

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    Homework Helper
    Gold Member

    What's U(x)? If the solution is given to you, then it must satisfy the differential equation.

    It'd be easier to spot errors if you post your work. Have you seen the LaTex tutorial for this forum yet? It's an easy way to post equations.
     
    Last edited: May 7, 2007
  4. May 7, 2007 #3
    i think you may think of the Fourier Transformation method
    and see it works or not~
     
  5. May 7, 2007 #4
    actually in the question paper it was given as v(x) instead U(x).
    Note,in the equation it was V(x),and in solution it is v(x) and is said that v(x) is also periodic as v(x+m)=v(x).

    I will try to adopt Latex notations.

    whose Fourier transform should I consider?
     
  6. May 7, 2007 #5
    um...i dont quite understand you question, sorry. (mainly confused with V(x), v(x) and U(x)... )
    would you mind telling me the in whole question once again..? :confused:
     
  7. May 7, 2007 #6
    OK,the question is I am to prove that=

    D^2f+[A+V(x)]f=0
    {A is a constant and V(x+m)=V(x),i.e.V(x) is periodic with period m}
    has the solutions of the form f=exp[ikx]v(x)where v(x+m)=v(x)
    i.e.v(x) is also periodic with period m
     
  8. May 7, 2007 #7
    After crude differentiation,I got this:
    f"=[v"(x)+2ikv'(x)]*exp[ikx]-k^2*f

    I write it as:
    f"(x)=-V(x)f(x)-k^2*f(x)
    and this can be written in desired form.

    Here I assume V(x)=-{U"+2ikU'}/U and A=-k^2.Right?

    Now I think we may check that V(x+m)=-{U"(x+m)+2ikU'(x+m)}/{U(x+m)}
    =-{U"+2ikU'}/U

    since after differentiation the period of the periodic function does not gets changed.Right?
     
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